Stationary Points & Turning Points (AQA GCSE Further Maths): Revision Note

Exam code: 8365

Roger B

Written by: Roger B

Reviewed by: Dan Finlay

Updated on

Stationary points & turning points

What are stationary points?

  • A stationary point is any point on a curve where the gradient is zero

  • To find stationary points of a curve

Step 1
Find the first derivative dydx

Step 2
Solve dydx=0 to find the x-coordinates of any stationary points

Step 3
Substitute those x-coordinates into the equation of the curve to find the corresponding y-coordinates

  • A stationary point may be either a local minimum, a local maximum, or a point of inflection 

Stat Point Illustr 1, A Level & AS Maths: Pure revision notes

Stationary points on quadratics

  • The graph of a quadratic function only has a single stationary point

  • For a positive quadratic this is the minimum; for a negative quadratic it is the maximum

    • No need to talk about 'local' here, as it is the overall minimum/maximum for the whole curve

Stationary Points min max for parabola illustr, A Level & AS Maths: Pure revision notes
  • The y value/coordinate of the stationary point is therefore the minimum or maximum value of the quadratic function

  • For quadratics especially, minimum and maximum points are often referred to as turning points

How do I determine the nature of stationary points on other curves?

  • For a graph there are two ways to determine the nature of its stationary points 

  • Method A

    • Compare the signs of the first derivative, dydx, (positive or negative) a little bit to either side of the stationary point

    • (After completing Steps 1 - 3 above to find the stationary points)

Step 4
For each stationary point find the values of the first derivative a little bit 'to the left' (i.e. for a slightly smaller x value) and a little bit 'to the right' (i.e. for a slightly larger x value) of the stationary point

Stat Points left right proviso, A Level & AS Maths: Pure revision notes

Step 5
Compare the signs (positive or negative) of the derivatives on the left and right of the stationary point

  • If the derivatives are negative on the left and positive on the right, the point is a local minimum (a u-shape)

  • If the derivatives are positive on the left and negative on the right, the point is a local maximum (an n-shape)

  • If the signs of the derivatives are the same on both sides (both positive or both negative) then the point is a point of inflection (ablank subscript bold divided by bold minus to the power of bold divided by shape)   

incr decr min max, A Level & AS Maths: Pure revision notes
Stationary Points point of inflection, A Level & AS Maths: Pure revision notes
  • Method B

    • Look at the sign of the second derivative (positive or negative) at the stationary point

    • (After completing Steps 1 - 3 above to find the stationary points)

Step 4
Find the second derivative d2ydx2

Step 5
For each stationary point find the value of d2ydx2 at the stationary point
i.e. substitute the x-coordinate of the stationary point into d2ydx2 and evaluate

  • If d2ydx2 is positive then the point is a local minimum

  • If d2ydx2 is negative then the point is a local maximum

  • If d2ydx2 is zero then the point could be a local minimum, a local maximum OR a point of inflection

    • use Method A to determine which

Examiner Tips and Tricks

  • Usually, using the second derivative (Method B above) is a much quicker way of determining the nature of a stationary point.

    • However, if the second derivative is zero it tells you nothing about the point

      • In this case you will have to use Method A

      • This method always works – see the Worked Example

Worked Example

Find the stationary points of

y=x3(3x220)

and determine the nature of each.

Start by expanding the brackets in the expression for y

y=3x520x3

Step 1
Find the first derivative

dydx=15x460x2

Step 2
Solve dydx=0

15x460x2=0

Divide through by 15

x44x2=0

Fully factorise, spotting the difference of two squares

x44x2=0x2(x24)=0

Solve to find the x-coordinates of the stationary (turning) points

x=0, 2, 2

Step 3
Find the corresponding y-coordinates

x=0,  y=(0)3(3(0)220)=0x=2, y=(2)3(3(2)220)=64x=2, y=(2)3(3(2)220)=64

Write the answers as coordinates, being careful to correctly match x's and y's

The stationary points are (0, 0), (-2, 64) and (2, -64)

To find the nature of each stationary point, start with Method B
Step 4
Find the second derivative

d2ydx2=60x3120x

Step 5
Evaluate the second derivative at each stationary point

x=0, d2ydx2=60(0)3120(0) = 0

x=2,  d2ydx2=60(2)3120(2)=240 < 0

x=2,  d2ydx2=60(2)3120(2)=240 > 0

The nature of two of the stationary points is now determined

(-2, 64) is a local maximum point
(2, -64) is a local minimum point

For the third stationary point, (0, 0) switch to Method A
Step 4
Compare first derivatives a little to the left and right of x=0

Choose x=1 and x=1

x=1, dydx=15(1)460(1)2=45 < 0

x=1,  dydx=15(1)460(1)2=45 < 0

Step 5
Interpret the result and state the answer

dydx has the same sign on both sides of (0, 0)

(0, 0) is a point of inflection

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Roger B

Author: Roger B

Expertise: Development Editor

Roger's teaching experience stretches all the way back to 1992, and in that time he has taught students at all levels between Year 7 and university undergraduate. Having conducted and published postgraduate research into the mathematical theory behind quantum computing, he is more than confident in dealing with mathematics at any level the exam boards might throw at you.

Dan Finlay

Reviewer: Dan Finlay

Expertise: Portfolio Lead

Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.